Difference between revisions of "009A Sample Midterm 2, Problem 5"

Revision as of 10:53, 9 April 2017

Find the derivatives of the following functions. Do not simplify.

(a) $f(x)=\tan ^{3}(7x^{2}+5)$

(b) $g(x)=\sin(\cos(e^{x}))$

(c) $h(x)={\frac {(5x^{2}+7x)^{3}}{\ln(x^{2}+1)}}$

Foundations:
1. Chain Rule
${\frac {d}{dx}}(f(g(x)))=f'(g(x))g'(x)$
2. Trig Derivatives
${\frac {d}{dx}}(\sin x)=\cos x,\quad {\frac {d}{dx}}(\cos x)=-\sin x$
3. Quotient Rule
${\frac {d}{dx}}{\bigg (}{\frac {f(x)}{g(x)}}{\bigg )}={\frac {g(x)f'(x)-f(x)g'(x)}{(g(x))^{2}}}$
4. Derivative of natural logarithm
${\frac {d}{dx}}(\ln x)={\frac {1}{x}}$

Solution:

(a)

Step 1:
First, we use the Chain Rule to get
$f'(x)=3\tan ^{2}(7x^{2}+5)(\tan(7x^{2}+5))'.$

Step 2:
Now, we use the Chain Rule again to get
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {3\tan ^{2}(7x^{2}+5)(\tan(7x^{2}+5))'}\\&&\\&=&\displaystyle {3\tan ^{2}(7x^{2}+5)\sec ^{2}(7x^{2}+5)(7x^{2}+5)'}\\&&\\&=&\displaystyle {3\tan ^{2}(7x^{2}+5)\sec ^{2}(7x^{2}+5)(14x).}\end{array}}$

(b)

Step 1:
First, we use the Chain Rule to get
$g'(x)=\cos(\cos(e^{x}))(\cos(e^{x}))'.$

Step 2:
Now, we use the Chain Rule again to get
${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {\cos(\cos(e^{x}))(\cos(e^{x}))'}\\&&\\&=&\displaystyle {\cos(\cos(e^{x}))(-\sin(e^{x}))(e^{x})'}\\&&\\&=&\displaystyle {\cos(\cos(e^{x}))(-\sin(e^{x}))(e^{x}).}\end{array}}$

(c)

Step 1:
First, we use the Quotient Rule to get
$h'(x)={\frac {\ln(x^{2}+1)((5x^{2}+7x)^{2})'-(5x^{2}+7x)^{2}(\ln(x^{2}+1))'}{(\ln(x^{2}+1))^{2}}}.$

Step 2:

Now, we use the Chain Rule to get

{\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {\frac {\ln(x^{2}+1)((5x^{2}+7x)^{2})'-(5x^{2}+7x)^{2}(\ln(x^{2}+1))'}{(\ln(x^{2}+1))^{2}}}\\&&\\&=&\displaystyle {\frac {\ln(x^{2}+1)2(5x^{2}+7x)(5x^{2}+7x)'-(5x^{2}+7x)^{2}{\frac {1}{x^{2}+1}}(x^{2}+1)'}{(\ln(x^{2}+1))^{2}}}\\&&\\&=&\displaystyle {{\frac {\ln(x^{2}+1)2(5x^{2}+7x)(10x+7)-(5x^{2}+7x)^{2}{\frac {1}{x^{2}+1}}(2x)}{(\ln(x^{2}+1))^{2}}}.}\end{array}}